Question

what is the angle between two vector if the ratio of their dot product and the magnitude of cross product is √3
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Answer 1

Answer: Let A and B are two vectors

dot product of A and B

A.B=|A||B|cosQ

cross product of Aand B

A*B=|A||B|sinQ

now ratio of these

A.B/A*B=|A||B|cosQ/|A||B|sinQ

>cosQ/sinQ=cotQ,

now according to your question ratio is equal to 3^1/2

cotQ=3^1/2

Q=30 degree

your answer is 30 degree

I Hope It Help You :-)